In known trapezoidal ABCD, AD / / BC, AC, BD intersect at O, passing through o as parallel line of ad intersects AB at m, intersects CD at n, and Mo = N0

In known trapezoidal ABCD, AD / / BC, AC, BD intersect at O, passing through o as parallel line of ad intersects AB at m, intersects CD at n, and Mo = N0

I'll do it``
In known trapezoidal ABCD, AD / / BC, AC, BD intersect at O, passing through o as parallel line of ad intersects AB at m, intersects CD at n, and Mo = N0
According to the picture of the title
It is easy to prove that the triangle AOD is similar to the triangle BOC
The picture shows that om: BC = Ao: OC
ON:BC=DO:OB
The triangle AOD is similar to the triangle BOC in that Ao: OC = do: ob
So Mo = N0